I did heed your advice and not look at your solution at all becauseJussi (2nd reply on this thread) had already pointed out the mistakein my observation. As soon as I saw his point, we solved the problemand got the answer. I then read your reply to see that we had arrivedat the same solution. Thank you for your generous help and caution.We grunt it out all the time, but missed a key observation that wouldhave alleviated the situation. Thank you for your time. Appreciated.

>On Tue, 15 Jan 2013 23:59:53 -0600, stony wrote:>> >> Hi,>> >> Need a little help with this. We are simplifying the following, but>> the solution is pretty lengthy and messy because of the enormous>> number of factors. I was thinking that may be I am missing seeing a>> pattern (some series or something). Is grunt work the only way to>> solve this or is there a pattern that can simplify the whole process?>> >> My daughter was trying to solve this, but ended up with the mess and>> then I got the same mess, but I thought there may be an easy way to>> simplify this that I may be missing.>> >> >> ((b-c)/((a-b)(a-c))) + ((c-a)/((b-c)(b-a))) + ((a-b)/((c-a)(c-b))) +>> (2/(b-a)) - (2/(c-a))>> >> >> of course, I took all the factors in the denominator and then started>> multiplying the numerator with the remaining factors to end up with a>>It would be nice if you had showed us your steps. You do recognize, >I hope, that a-b and b-a are just the same thing with a factor of -1 >pulled out? So if you're treating (a-b) and (b-a) as different >factors in finding your common denominator, you're doing much more >work than you need to. In fact there are only three factors, a-b, a->c, b-c, so your common denominator will be (a-b)(a-c)(b-c).>>PLEASE, with that hint, solve it on your own, and only ten look at >what I've done.>>>>>>>>>>>>>>>>First step: write all the binomial factors in the same form:>>(b-c)/[(a-b)(a-c)] + (-1)(a-c)/[(b-c)(-1)(a-b)] > + (a-b)/[(-1)(a-c)(-1)(b-c)] + 2/[(-1)(a-b)] - 2/[(-1)(a-c)]>>(b-c)/[(a-b)(a-c)] + (a-c)/[(a-b)(b-c) > + (a-b)/[(a-c)(b-c)] + (-2)/(a-b) + 2/(a-c)>>(b-c)^2/[(a-b)(a-c)(b-c)] + (a-c)^2/[a-b)(a-c)(b-c)]> + (a-b)^2/[(a-b)(a-c)(b-c)] > + (-2)(a-c)(b-c)/[(a-b)(a-c)(b-c)] > + 2(a-b)(b-c)/[(a-b)(a-c)(b-c)]>>Temporarily disregarding the common denominator, you have a numerator >of>> b^2 -2bc + C^2> +a^2 -2ac + c^2> +a^2 -2ab + b^2 > -2ab +2ac +2bc -2c^2> +2ab -2b^2 -2ac +2bc>>Which is>>2a^2 - 2ab - 2ac + 2bc>= 2a(a-b) -2c(a-b)>= 2(a-b)(a-c)>>Now restoring the denominator, the whole fraction is>>2(a-b)(a-c)/[(a-b)(a-c)(b-c)] = 2/(b-c)